Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{4x - 28}{x + 3} \div \dfrac{x^2 - 14x + 49}{-4x + 28} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{4x - 28}{x + 3} \times \dfrac{-4x + 28}{x^2 - 14x + 49} $ First factor the quadratic. $q = \dfrac{4x - 28}{x + 3} \times \dfrac{-4x + 28}{(x - 7)(x - 7)} $ Then factor out any other terms. $q = \dfrac{4(x - 7)}{x + 3} \times \dfrac{-4(x - 7)}{(x - 7)(x - 7)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ 4(x - 7) \times -4(x - 7) } { (x + 3) \times (x - 7)(x - 7) } $ $q = \dfrac{ -16(x - 7)(x - 7)}{ (x + 3)(x - 7)(x - 7)} $ Notice that $(x - 7)$ appears twice in both the numerator and denominator so we can cancel them. $q = \dfrac{ -16\cancel{(x - 7)}(x - 7)}{ (x + 3)\cancel{(x - 7)}(x - 7)} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $q = \dfrac{ -16\cancel{(x - 7)}\cancel{(x - 7)}}{ (x + 3)\cancel{(x - 7)}\cancel{(x - 7)}} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $q = \dfrac{-16}{x + 3} ; \space x \neq 7 $